Math 308 Week in Review 13 — Solutions

The equation is in standard linear form $y' + p(t)\,y = g(t)$ with $p(t) = 2$, so the integrating factor is

Multiplying both sides by $e^{2t}$ collapses the left side into a single derivative:

Integrating gives $e^{2t}\,y = e^{t} + C$, so $y(t) = e^{-t} + C\,e^{-2t}$. The initial condition forces $y(0) = 1 + C = -1$, hence $C = -2$.

Check: $y' = -e^{-t} + 4e^{-2t}$, so $y' + 2y = -e^{-t} + 4e^{-2t} + 2e^{-t} - 4e^{-2t} = e^{-t}$. $\checkmark\;$ Also $y(0) = 1 - 2 = -1$. $\checkmark$

Structural observation. The general solution decomposes as $y = \underbrace{e^{-t}}_{y_p} + \underbrace{C\,e^{-2t}}_{y_h}$. The particular piece $e^{-t}$ comes from the integrating-factor calculation; the homogeneous piece $C e^{-2t}$ encodes the response of $y' + 2y = 0$ on its own. The constant $C$ is fixed by the initial condition.

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Dividing by $y^2$ (valid since $y(0) = 1 \neq 0$):

Multiplying by $-1$ and rearranging gives $\dfrac{1}{y} = x^2 - C$. The initial condition $y(0) = 1$ forces $1 = -C$, so $C = -1$ and $\dfrac{1}{y} = x^2 + 1$:

Check: $y' = -\dfrac{2x}{(1+x^2)^2}$ and $-2xy^2 = -\dfrac{2x}{(1+x^2)^2}$. $\checkmark\;$ Also $y(0) = 1$. $\checkmark$

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The Existence and Uniqueness theorem (Sec. 3.2) requires the coefficients $p(t)$, $q(t)$, and the right-hand side $g(t)$ in the standard form $y'' + p(t)\,y' + q(t)\,y = g(t)$ to all be continuous on an open interval containing the initial point. Dividing through by $(t^2 - 9)$ to put the equation in standard form:

The points where $p$, $q$, or $g$ fails to be continuous:

• $t^2 - 9 = 0 \;\Rightarrow\; t = \pm 3$  (from $p$, $q$, and $g$).
• $t - 1 = 0 \;\Rightarrow\; t = 1$  (from $q$).
• $\sqrt{8-t}$ is real-valued only when $8 - t \geq 0$, i.e., $t \leq 8$, so $g$ is undefined for $t > 8$.

Marking these on the number line:

The point $t_0 = 2$ lies between $1$ and $3$. The largest open interval around $2$ that avoids both $1$ and $3$ (and stays below $8$) is therefore

Why open? The theorem requires an open interval because continuity is needed on a neighborhood of every interior point; a closed endpoint at a singularity would violate that.

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Setting $f(y) = (y-1)(y-3)^2 = 0$ gives equilibria $y = 1$ and $y = 3$. Note that $y = 3$ is a double root of $f$ — a hint that something unusual happens there.

For sign analysis, observe that $(y-3)^2 \geq 0$ everywhere and is zero only at $y = 3$. So the sign of $f(y)$ is determined by the sign of $(y-1)$ everywhere except at the equilibria themselves:

• $y < 1$:   $(y-1) < 0$, $(y-3)^2 > 0$  $\Rightarrow$  $f(y) < 0$, solutions decrease.
• $1 < y < 3$:   $(y-1) > 0$, $(y-3)^2 > 0$  $\Rightarrow$  $f(y) > 0$, solutions increase.
• $y > 3$:   $(y-1) > 0$, $(y-3)^2 > 0$  $\Rightarrow$  $f(y) > 0$, solutions increase.

The phase line, drawn as a horizontal $y$-axis with arrows indicating direction of motion:

For the direction field, slopes are constant along horizontal lines (autonomous equation), and sample slopes are easy to read off:

• $y = 0$: $f = (-1)(9) = -9$  (very steep downward)
• $y = 1$: $f = 0$  (horizontal — equilibrium)
• $y = 2$: $f = (1)(1) = 1$  (modest upward)
• $y = 3$: $f = 0$  (horizontal — equilibrium)
• $y = 4$: $f = (3)(1) = 3$  (modest upward)
• $y = 5$: $f = (4)(4) = 16$  (very steep upward)

Classifying each equilibrium: at $y = 1$, trajectories with $y < 1$ decrease further (moving away from $y=1$), and trajectories with $y > 1$ increase (also moving away) — both sides flee, so $y = 1$ is unstable. At $y = 3$, trajectories in the strip $1 < y < 3$ increase toward $3$, but trajectories with $y > 3$ also increase, moving away from $3$. So $y = 3$ is semi-stable — attracting from the $y < 3$ side, repelling from the $y > 3$ side. This one-sided behavior is the signature of an even-order zero (here the squared factor $(y-3)^2$).

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Let $Q(t)$ denote kilograms of salt at time $t$ (minutes). Since the volume is constant at $50\,\text{L}$ (in-rate equals out-rate), the concentration in the tank is $\dfrac{Q}{50}\,\text{kg/L}$. The rate equation is

which rearranges to the linear IVP

The integrating factor is $\mu(t) = e^{2t/25}$, giving

The initial condition $Q(0) = 25 + C = 0$ gives $C = -25$, so

As $t \to \infty$, $Q(t) \to 25\,\text{kg}$. Physically, the tank's concentration approaches the inflow concentration $0.5\,\text{kg/L}$, and at that concentration the $50\,\text{L}$ tank holds $0.5 \times 50 = 25\,\text{kg}$ of salt. The system reaches equilibrium when input and output rates of salt balance.

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For each part, read off the characteristic roots from the form of $y(t)$, then build the characteristic polynomial $(r-r_1)(r-r_2)$ and translate it back to an ODE.

Part (a)  $y(t) = (1 + 3t)\,e^{2t} = 1\cdot e^{2t} + 3t\cdot e^{2t}$.

Both basic solutions $e^{2t}$ and $te^{2t}$ are present, which is the signature of a repeated root $r = 2$. So the characteristic equation is

and the ODE is therefore $y'' - 4y' + 4y = 0$.

Initial conditions. $y(0) = (1 + 0)\cdot 1 = 1$. Differentiating:

Part (b)  $y(t) = e^{-t}\bigl(2\cos(2t) - 3\sin(2t)\bigr)$.

The factor $e^{-t}$ paired with $\cos(2t), \sin(2t)$ signals complex roots $r = -1 \pm 2i$. So the characteristic polynomial is

and the ODE is $y'' + 2y' + 5y = 0$.

Initial conditions. $y(0) = 1\cdot(2 - 0) = 2$. Differentiating with the product rule:

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The characteristic equation $r^2 - 3r + 2 = (r-1)(r-2) = 0$ has roots $r_1 = 1$ and $r_2 = 2$, so the homogeneous solutions are $e^t$ and $e^{2t}$. By superposition, write $y_p = y_{p,1} + y_{p,2}$, one piece for each forcing term.

For the $4 e^{t}$ piece: the trial form $A e^t$ duplicates the homogeneous solution $e^t$ (since $r = 1$ is a hom root), so multiply by $t$:

Substituting:

so $A = -4$ and $y_{p,1} = -4\,t\,e^{t}$.

For the constant $5$ piece: try $y_{p,2} = B$ (constant); then $y_{p,2}' = y_{p,2}'' = 0$, and the equation becomes $0 - 0 + 2B = 5$, giving $B = \tfrac{5}{2}$.

Why no resonance for the constant? A constant forcing corresponds to forcing exponent $r = 0$, which is not a root of $r^2 - 3r + 2$. The constant ansatz works without modification; only the $e^t$ piece resonates.

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Part (a): Find values of $m$. Substitute $y = t^m$, $y' = m\,t^{m-1}$, $y'' = m(m-1)t^{m-2}$ into the homogeneous equation $t^2 y'' - 3t y' + 3y = 0$:

The bracket is the indicial equation:

Part (b): Wronskian / linear independence.

For $t > 0$, $W = 2t^3 \neq 0$, so $\{y_1, y_2\} = \{t,\,t^3\}$ is a fundamental set of solutions.

Part (c): Variation of parameters. First put the equation in standard form (coefficient of $y''$ equals 1) by dividing through by $t^2$:

The variation-of-parameters formula gives

Compute each integrand:

Integrate:

Therefore

Check: $y_p' = -10t$, $y_p'' = -10$. Then $t^2(-10) - 3t(-10t) + 3(-5t^2) = -10t^2 + 30t^2 - 15t^2 = 5t^2.$ $\checkmark$

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Part (a): Free damped motion. Compute the discriminant first:

The characteristic roots are

So $u(t) = e^{-t}\bigl(c_1\cos t + c_2\sin t\bigr)$. Apply $u(0) = 1$: $c_1 = 1$. Differentiate:

Part (b): IVP with impulse + step. A hammer impulse of magnitude $3\,\text{N}\!\cdot\!\text{s}$ at $t = 1$ becomes the term $3\,\delta(t-1)$ on the right-hand side. A constant force of $5\,\text{N}$ that switches on at $t = 4$ becomes $5\,u_4(t)$. Starting from rest at equilibrium gives both initial conditions zero:

Part (c): Steady-state for $F(t) = -10\cos(2t)$. Try $u_p = A\cos(2t) + B\sin(2t)$. Differentiating:

Substitute into $u_p'' + 2u_p' + 2u_p$ and group by $\cos(2t)$ and $\sin(2t)$:

Match coefficients with $-10\cos(2t) + 0\sin(2t)$:

From the second equation, $B = -2A$. Substitute into the first: $-2A + 4(-2A) = -10A = -10$, so $A = 1$ and $B = -2$.

Amplitude / phase form $u_p = R\cos(2t + \delta)$. Expanding:

Matching: $R\cos\delta = 1$ and $-R\sin\delta = -2$ (so $R\sin\delta = 2$). Therefore

Both $\cos\delta > 0$ and $\sin\delta > 0$, so $\delta \in (0, \pi/2)$: $\delta = \arctan 2 \approx 1.107\,\text{rad} \approx 63.4^\circ$.

Why is the steady state the same as $u_p$? Because the system is damped ($\gamma > 0$), every homogeneous solution decays as $t \to \infty$, leaving only the bounded particular solution. So $u_{\rm steady-state}(t) = u_p(t)$ — no need to solve for $c_1, c_2$.

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Splitting the integral at the breakpoint $t = 2$ and the two pieces apart:

The constant piece is straightforward:

For the $t^2$ piece, two applications of integration by parts give the antiderivative

(differentiating the right side recovers the integrand). Evaluating from $t = 0$ to $t = 2$:

Adding the two pieces, the $\dfrac{4e^{-2s}}{s}$ terms cancel exactly:

Grouping the $1/s^3$ terms:

Cross-check via the unit-step formula (not required, but reassuring). Write $f(t) = t^2 + (4 - t^2)\,u_2(t)$. Expanding $4 - t^2$ around $t = 2$: $4 - t^2 = -[(t-2)^2 + 4(t-2)]$. By the second shift theorem,

Adding $\mathcal{L}\{t^2\} = \dfrac{2}{s^3}$ recovers the same expression. $\checkmark$

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Let $Y = \mathcal{L}\{y\}$. Taking the Laplace transform of both sides and using $\mathcal{L}\{y'\} = sY - y(0)$ and $\mathcal{L}\{y''\} = s^2 Y - s\,y(0) - y'(0)$:

Collecting:

Completing the square in the denominator: $s^2 + 4s + 5 = (s + 2)^2 + 1$. To match the inverse Laplace pairs centered at $s = -2$, rewrite the numerator as $2s + 7 = 2(s + 2) + 3$:

Inverting using $e^{at}\cos(bt) \leftrightarrow \dfrac{s-a}{(s-a)^2 + b^2}$ and $e^{at}\sin(bt) \leftrightarrow \dfrac{b}{(s-a)^2 + b^2}$ with $a = -2$ and $b = 1$:

Check: $y(0) = 1\cdot(2 + 0) = 2$. $\checkmark\;$ Computing $y'(0)$ via the product rule confirms $y'(0) = -1$. $\checkmark$

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Taking the Laplace transform with both initial values zero and $\mathcal{L}\{u_2(t)\} = e^{-2s}/s$:

Partial fractions on $\dfrac{1}{s(s^2+1)}$:

Clearing denominators gives $1 = A(s^2 + 1) + (Bs + C)\,s$. At $s = 0$: $A = 1$. Comparing $s^2$ coefficients: $0 = A + B$, so $B = -1$. Comparing $s^1$ coefficients: $C = 0$. Therefore

Inverting (without the shift for now): let $f(t) = \mathcal{L}^{-1}\!\left\{\dfrac{1}{s} - \dfrac{s}{s^2+1}\right\} = 1 - \cos t.$ The second shift theorem then gives $y(t) = u_2(t)\,f(t-2)$:

Sanity check: at $t = 2^+$, $y(2) = 1 - \cos(0) = 0$ and $y'(2) = \sin(0) = 0$, so the solution is continuous and $C^1$ at the activation point — as expected for a step-forced (not impulse-forced) IVP.

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$P(x) = x$ and $Q(x) = 1$ are polynomials, so $x = 0$ is an ordinary point and the series converges for all $x$ (radius $R = \infty$).

Substituting $y = \displaystyle\sum_{n=0}^\infty a_n x^n$ and re-indexing each term to a common $x^n$ form:

Substituting into $y'' + xy' + y = 0$:

Setting each coefficient to zero and dividing by $(n+1)$ gives the recurrence

The even-indexed coefficients propagate from $a_0$ and the odd-indexed from $a_1$. For the even chain:

For the odd chain:

Taking $(a_0, a_1) = (1, 0)$ and $(0, 1)$ in turn produces the basis solutions to order $x^4$:

(The next nonzero term in $y_2$ is $\tfrac{x^5}{15}$, beyond the $x^4$ truncation.)

Tip: $y_1(x)$ here is in fact the Maclaurin series of $e^{-x^2/2}$ — you can verify $\frac{d}{dx}e^{-x^2/2} = -x\,e^{-x^2/2}$ and that $y_1$ satisfies the ODE in closed form.

Applying the initial conditions: $y(0) = a_0 = 2$ and $y'(0) = a_1 = -1$, so $y(x) = 2\,y_1(x) - y_2(x)$. Up to $x^4$:

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Part (a): Eigenvalues of $A$.

So $\lambda = 2$ is a repeated eigenvalue with algebraic multiplicity 2.

Eigenspace check. Compute $A - 2I = \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}$, which has rank 1. So the geometric multiplicity is $2 - 1 = 1$ — only one independent eigenvector. The matrix is defective; we'll need a generalized eigenvector for the second hom solution.

Eigenvector and generalized eigenvector. $(A-2I)\mathbf{v} = \mathbf{0}$ gives $v_1 - v_2 = 0$, so $\mathbf{v} = \binom{1}{1}$. Then $(A-2I)\mathbf{u} = \mathbf{v}$ gives $u_1 - u_2 = 1$; choose $\mathbf{u} = \binom{1}{0}$.

Homogeneous solutions.

Part (b): Particular solution by superposition. The forcing has two pieces, so build $\mathbf{x}_p = \mathbf{x}_{p,1} + \mathbf{x}_{p,2}$.

Piece 1: exponential $\binom{2}{0}e^{t}$ with exponent $\mu = 1$. Since $\mu = 1$ is not an eigenvalue (the eigenvalue is $2$), there is no resonance. Try $\mathbf{x}_{p,1} = \mathbf{a}\,e^{t}$. Substituting into $\mathbf{x}' = A\mathbf{x} + \binom{2}{0}e^{t}$:

Compute $I - A = \begin{pmatrix} -2 & 1 \\ -1 & 0 \end{pmatrix}$ and solve:

So $\mathbf{x}_{p,1}(t) = \binom{0}{2}e^{t}$.

Piece 2: constant $\binom{0}{4}$. Try $\mathbf{x}_{p,2} = \mathbf{c}$ (constant), giving $\mathbf{0} = A\mathbf{c} + \binom{0}{4}$, i.e., $A\mathbf{c} = \binom{0}{-4}$.

From the first, $c_2 = 3c_1$; sub into the second: $4c_1 = -4 \Rightarrow c_1 = -1$, $c_2 = -3$. So $\mathbf{x}_{p,2} = \binom{-1}{-3}$.

Combine.

Part (c): General solution. $\mathbf{x} = c_1\,\mathbf{x}_1 + c_2\,\mathbf{x}_2 + \mathbf{x}_p$:

Verification of $\mathbf{x}_p$. $\mathbf{x}_p' = \binom{0}{2e^t}$. Then $A\mathbf{x}_p = \begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix}\!\binom{-1}{2e^t - 3} = \binom{-3 - 2e^t + 3}{-1 + 2e^t - 3} = \binom{-2e^t}{2e^t - 4}$. Adding $\mathbf{g}(t) = \binom{2e^t}{4}$ gives $A\mathbf{x}_p + \mathbf{g} = \binom{0}{2e^t} = \mathbf{x}_p'.$ $\checkmark$

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